Additive inverse of a number vs multiplying by $-1$

Question

To give the context, I've been trying to look at different ways to convince myself how $-\times - = +$

Additive inverse of $a$ is written as $-a$

As an example the additive inverse of $-3$ is written as $-(-3)$
Also $-1$ times $-3$ is written as $(-1)\times (-3)$

Both above expressions evaluate to the same quantity $3$.
I guess it is easy to see why the additive inverse of $-3$ equals $3$ simply by staring at the equation $3+(-3) = 0$

However it must be very difficult to convince oneself why the second expression $(-1)\times (-3)$ evaluates to $3$ too. Both these operations seem related. I'm trying to figure out connection/intuition behind taking additive inverses and multiplying by $-1$. Help is appreciated. Thanks!

Answer 1

The standard proof goes along the lines of $$ 0\times(- 3)=0\\ (1+(-1))\times(-3)=0\\ 1\times (-3)+(-1)\times(-3)=0\\ (-3)+(-1)\times(-3)=0 $$ and we see that $(-1)\times(-3)$ is an (the) additive inverse of $(-3)$.

Answer 2

A way to provide insight that additive inverses add up to $0$ is:

You are doing a certain distance in a direction in a coordinate plane. Let that distance be $x$ units. You are going the same distance in the opposite direction, or going to where you initially were. We moved the same distance, $x$ both times, but you were moving "backwards" $x$ the second time, which can be denoted by $-(x)$. Since you end up at your original location, you can claim that $x+(-x)=0$.

Now, we know additive inverses add up to $0$. Now for the multiplication of two negatives.

Using the information that additive inverses add up to $0$, define $x=ab+(-a)(b)+(-a)(-b)$, where $a$ and $b$ are real numbers.

$$x=ab+(-a)(b)+(-a)(-b)\implies x=ab+(-a)[(b)+(-b)]\implies x=ab+(-a)[0]$$

Which means $x=ab$. Also:

$$x=ab+(-a)(b)+(-a)(-b)\implies x=b[(a)+(-a)]+(-a)(-b)\implies x=b[0]+(-a)(-b)$$

Which means $x=(-a)(-b)$. Since $x=ab$ and $x=(-a)(-b)$, what can you tell?